# The continuous measure function

If $f$ is a Lebesgue measurable function on $R^n$, we define $$K(t)=\lambda\{x\in R^n:|f(x)|>t\}.$$ I want to prove that

1. $\int_0^{\infty}K(t)dt=\int_{R^n} |f(x)|dx$

2. If $f\in L^1 (R^n)$, then $$\lim_{s\rightarrow t^-} K(s)=K(t).$$

I have no idea about these two problems.

-

## 1 Answer

I will give hints:

1. Use Fubini theorem for non-negative functions and $F(x,t):=\chi_S(x,t)$, where $S:=\{(x,t)\in\Bbb R^2,|f(x)|>t\}$.

2. If $f$ is integrable, then $\{K(s)\}_s$ is decreasing and consists of sets of finite measure. This works, as noted in the comments, if $K(t):=\lambda\{x,|f(x)|\color{red}{\geqslant} t\}$. Otherwise, we take $n=1$, $f=\chi_{(0,1)}$ and $t=1$.

-
How to deal with f(x)=t for the second problem. –  89085731 Dec 7 '12 at 22:31
A conditional statement is true, provided the hypothesis is false. –  GEdgar Dec 7 '12 at 22:33
Good point, for example when $n=1$, $f=\chi_{(0,1)}$ and $t=1$, then $K(1)=0$ but $K(s)=1$ for $s<t$. But if we define $K(t):=\lambda\{x, f(x)\geqslant t\}$ it will work. –  Davide Giraudo Dec 7 '12 at 22:36
so the problem is wrong,right? –  89085731 Dec 7 '12 at 22:42
It seems (or there is just a typo). –  Davide Giraudo Dec 7 '12 at 23:30